B Calculating Orbital Radius for Hydrogen Atom with Given Angular Velocity

AI Thread Summary
The discussion revolves around calculating the orbital radius of a hydrogen atom using classical mechanics, specifically with an electron's angular velocity set at 10^16 s^-1. The initial calculations equate centripetal and electrostatic forces, leading to a derived radius that differs from the known Bohr radius. Participants clarify that classical mechanics cannot accurately describe electron behavior, as quantum mechanics dictates that electrons do not follow fixed orbits. The calculations may yield a valid radius based on the chosen velocity, but this does not align with quantum mechanical principles. Ultimately, the derived radius reflects classical assumptions rather than the true nature of electron orbits in hydrogen.
psy
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Hey guys,

The following thing got me a little bit messed up.

I want to calculate the orbital radii of an Hydrogen atom if the angular velocity of the electron is 10^16 * s^-1.

At first i set the centripetal force and the electrostatic force as equals.

( m * v^2 )/ r =k * (e^2) / r^2

v = ω * r

( m * ω^2 * r^2) / r = k * (e^2) / r^2

r ^3 = k * (e^2) / ( m * ω^2 )

r^3 = 8,99 * 10^9 Nm^2 / C * (1,60 * 10^-19 C )^2 / 9.11 10^-31 Kg * (10^16 s^-1 )^2

r^3 = (23,015 * 10^-29 Nm^2 C ) / 91,1 Kg/s^2

r =13,61 * 10^-11

While the Hydrogen radii is 0,52 * 10^-10 , I am checking it over and over again and can't find where i messed up.
Can someone tell me where i was wrong with the calculation?

Kind Regards
 
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psy said:
Can someone tell me where i was wrong with the calculation?
You are using classical mechanics, and the electron obeys the laws of quantum mechanics not classical mechanics.

In quantum mechanics the electron has no definite speed or position and does not follow a circular path around the nucleus the way the planets orbit the sun; instead we just have some probability of finding it in some spot near the nucleus if we look for it there. That ##5.2\times{10}^{-11}## radius is where that probability peaks.

(Googling for "hydrogen radius" will find many more links and the wikipedia article is not bad, but the math may quickly move past what belongs in a B-level thread).
 
Hi, I am coming to this cold but where does your initial figure for the angular velocity of the electron come from? This chosen value has to be right to get the right orbit radius.
Your starting equation looks ok (if k is given the right value) for a classical orbit. This Hyperphysics link starts in much the same way but the process is in terms of Energy. Start from a different direction, perhaps? (With QM in mind)
 
In the exercise I am doing its supposed that the electron moves in a circular orbit around the proton with the given velocity,
so I am tryng the classical mechanics with the centripetal and electrostatic force,where i can plug in the velocity of 10^16 s^-1 .

As the k I used 1/ 4*π*ε = 8,99 * 10^9 Nm/C^2 .
 
Why would you expect to get the value of the first Bohr radius? You get the radius corresponding to that velocity, in the classical mechanics framework.
It may be the right answer even it may irrelevant from the point of view of QM model. Even in the semi-classical Bohr model, there are more than one possible values for the radius of the orbit.
 
psy said:
In the exercise I am doing its supposed that the electron moves in a circular orbit around the proton with the given velocity,
so I am tryng the classical mechanics with the centripetal and electrostatic force,where i can plug in the velocity of 10^16 s^-1 .

As the k I used 1/ 4*π*ε = 8,99 * 10^9 Nm/C^2 .
So don't worry. You got an answer but it's not surprising its the Bohr radius because it makes different assumptions.
 
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